Question: What value of $x$ will give the maximum value for $-x^2- 6x + 12$?
Solution: We start by completing the square: \begin{align*}
-x^2 -6x +12 &= -(x^2 + 6x) + 12\\ &= -(x^2 + 6x + (6/2)^2 - (6/2)^2) + 12\\ &= -((x+3)^2 -3^2) + 12 \\&= -(x+3)^2 +3^2 + 12 \\&= -(x+3)^2 + 21.\end{align*}Since the square of a real number is at least 0, we have $(x+3)^2\ge 0$, so $-(x+3)^2 \le 0$.  Therefore, $-(x+3)^2 + 21$ is at most 21.  Since $(x+3)^2 =0$ when $x=-3$, this maximum of $21$ is achieved when $x= \boxed{-3}$.